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3n+k 2 if n 1 mod2 Fk(n)= n ≡ (1) ( if n 0 mod2 2 ≡ Here, k is an odd positive integer. For the rest of the paper, unless otherwise specified, whenever we refer to k, we mean odd positive integer. Generalized collatz functions have been studied extensively in the past [6] and our particular extension has been studied in [5] [4] [8] [1]. It has been known that such generalized collatz functions yield multiple cycles and are detailed in some previous studies [5]. We present a closed form formula for such cycles and the implications thereof. In the concluding section we show that such cycles yield solution to the Diophantine equations of the form 2n 3m = k. − 2 Definitions

We define the terms that will be used often in this paper. Path : Path of F (n) is the set of numbers that are reached during successive iter- • k ations of the function. Path of F (n) = F (n), F 2(n), F 3(n), . For example, k { k k k ···} for k = 5 and n = 12, the subsequent numbers are 12 6 3 7 13 → → → → →

2 22 11 19 31 49 76 38 19 31 . Therefore the path of → → → → → → → → → ··· F (12) is 12, 6, 3, 7, 13, 22, 11, 19, 31, 49, 76, 38, 19, 31 and 12, 6.. are elements of 5 { } the path.

Cycle : A cycle in F is the list of values beginning from the minimum number • k that are reached repeatedly in the path of F (n) for some n N. We will refer to k ∈ the minimum number in the cycle as T0 in the rest of the paper. Thus, a cycle is T , F (T ), F 2(T ), F 3(T ), , F s(T ) with F s+1(T )= T and s is the number { 0 k 0 k 0 k 0 · · · k 0 } k 0 0 of steps. For example, for k = 5 and n = 12 the path is 12 6 3 7 13 { → → → → → 22 11 19 31 49 76 38 19 31 49 76 38 19, . → → → → → → → → → → → → ···} The cycle is 19, 31, 49, 76, 38 . { } Set of Cycles ζ : We define ζ as a set of the minimum values of all the cycles • k k in F . For example ζ 1, 5, 19, 23, 187, 347 as we have 6 known cycles in F . k 5 ⊇ { } 5 3n+k up iteration : An up step is the usage of 2 once, n being odd. An up iteration • 3n+k is the successive application of 2 due to a series of odd numbers in the path ending in an even number. The number of steps applied in the entire up iteration is represented by u throught this paper.

n down iteration : A down step is the usage of 2 once, n being even. A down • n iteration is the successive application of 2 due to a series of even numbers in the path. The number of steps applied in the entire down iteration is represented by d throughout this paper.

orb : An orb is defined as one up iteration and one down iteration. • s-cycle : A s-cycle is a set of s orbs starting with an up iteration such that ending • number is same as the starting number. For the rest of the paper we will represent th a s-cycle with s orbs as u1, d1, u2, d2, , us, ds . The i up iteration step count { th · · · } is represented by ui, while i down iteration step count is represented by di. We will also interchangeably use u , d to denote the full s-cycle with s orbs. { i i} Original cycle of F : A cycle u , d is an original cycle of F if k is the smallest • k { i i} k number where cycle appears in Fk sequence. Cycle inheritance: A cycle is said to be inherited from F if there exists a cycle • r in F and F with identical sequence of u , d where k > r. k r { i i}

3 We would use the following definitions consistently throughout the paper • s U = ui 1 Xs D = di 1 X − Pi 1 u +d u u Ps u α = 2 1 j j (3 i 2 i )3 i+1 j i − s α = αi 1 X β = 2U+D 3U −

Note that exponent of 2 in α1 and 3 in αs is zero.

3 Collatz Sequence

Collatz sequence is perhaps the simplest to describe unsolved problem in mathematics. There are two parts to the collatz conjecture

There is no non-trivial cycle in a collatz sequence. • Starting with any number, collatz sequence always converges. • These two statements can be integrated into a single statement and the collatz conjecture can be: for all positive integers, there exists an r such that r steps of applying collatz function yields 1, or n r [f r(n) = 1] ∀ ∃ Both sub-conjectures are unproven and so is the collatz conjecture. Wikipidea (https://en.wikipedia.org/wiki/Collatz conjecture) page on collatz sequence and Lagaris (2010) [9] provide an excellent overview of the prob- lem, previous research, current status, and progress towards solution. Eliahou, Shalom [2] proved that the length of the non trivial cycle follows

p = 301994 a + 17087915 b + 85137581 c where a, b, c are non negative integer, b 1, ac = 0. Per the data available in 1993, the ≥ minimum length of non trivial cycle was calculated to 1708791. The latest calculations based on computation till 87 260 increases minimum cycle length to 10439860591. × Steiner (1977) [12] proved that the only 1-orb cycle is the trivial cycle. This approach was used by Simons (2004) and subsequently by Simon de Wegner (2005) [11] [10] that there is no non-trivial cycle with less than 69 orbs. As more sequences with larger n are computed, lower limit on the orbs, the minimum number of elements, and the minimum

4 number in the non trivial cycle will continue to grow. However, that is not the proof that no non trivial cycle exists. Lagaris (1985) [7] makes a probabilistic heuristic argument that collatz sequence decreases in long run. The argument is based on considering each step as being odd or even with equal probability. Thus, one can show that the average size decrease between 3 two consecutive odd numbers by a factor of 4 . Therefore, one can argue that on an average collatz sequence converges over long run. One of the issue with this argument is collatz numbers are not uncorrelated random number. Furthermore, even if one ignores this shortcoming, it still does not prove that collatz sequence starting with every number will converge. Terrance Tao (2019) [13] posted a note proving that collatz conjecture is ”almost” true for ”almost” all numbers. This result is being considered as the most significant advancement towards collatz conjecture in last two decades. Though, this still does not prove the conjecture. Collatz conjecture is generally considered a very hard problem to solve. Lagaris [9] notes

”The 3x + 1 problem remains unsolved, and a solution remains unapproachable at present.”

Paul Erdos famously commented:

”Mathematics is not yet ready for such problem”.

4 Cycles in Generalized Collatz Sequence

In this section we derive the formula for a s-cycle in Fk through a set of Theorems.

Theorem 1 : Each cycle in Fk for a given k can be uniquely represented by its lowest element. Proof: Let each number be a node on a graph, then each node can only be directed towards one point, therefore the successive number for each node is fixed. Since there is only one path originating from a node, if a number is the lowest element of a cycle, it uniquely describes the cycle. Therefore, all information about the cycles of Fk is cap- tured by ζk.

Theorem 2 : Let odd number T0 be the starting point of an orb. The output of Fk(n) after an up iteration with u steps is 3uT + k(3u 2u) O = 0 − . u 2u

Proof : Let the output be O1 after one up step. Since T0 is odd 3T + k O = 0 1 2

5 Therefore,the formula holds for u = 1. Let the formula be true for u, then 3uT + k(3u 2u) O = 0 − u 2u For u +1 up steps

u u u 3 T0+k(3 −2 ) 3 u + k O = 2 u+1 2 3u+1T + k(3u+1 2u 3 + 2u) = 0 − · 2u+1 3u+1T + k(3u+1 2u+1) = 0 − 2u+1

Therefore, the formula is true for u+1 up steps. By mathematical induction, the formula is true for all u N. ∈

Theorem 3: Let T0 be the starting point of an orb. The output of Fk(T0) after s orbs with sequence u , d , u , d , , u , d is { 1 1 2 2 · · · s s} 3U T + kα T = 0 (2) s 2U+D

Proof : Let Tr be the output of Fk after r orbs. From Theorem 2, output after u1 up steps 3u1 T + k(3u1 2u1 ) = 0 − 2u1 st and, the output T1 after 1 orb - u1 up steps followed by d1 down steps is u u u 3 1 T0+k(3 1 −2 1 ) 2u1 T1 = 2d1 3u1 T + k(3u1 2u1 ) = 0 − 2u1+d1 therefore, the formula holds for one orb. Let the formula be true for a m orbs, therefore 3U T + kα T = 0 m 2U+D for m +1 orbs

i−1 P u +d Pm u 3P ui T +k(Pm (2 j=1 j j )(3ui −2ui )(3 j=i+1 j )) 3um 0 i=1 + k(3um+1 2um+1 ) 2P ui+di Tm+1 = − 2um+1+dm+1 − m+1 Pi 1 u +d Pm+1 u 3P ui T + k( (2 j=1 j j )(3ui 2ui )(3 j=i+1 j )) = 0 i=1 − 2P ui+di P

6 therefore, the formula is true for m +1 orbs if it is true for m orbs. By mathematical induction, the formula is true for all m N. ∈

Theorem 4 : A trivial cycle k 2k exists for all Fk. { → } 3k+k Proof : Since k is odd, path for Fk(k) is 2 = 2k. Since 2k is even, Fk(2k) = 2k = k. Therefore, k 2k k is always a cycle in F . 2 → → k Theorem 5 : Minimal point T and the orbs u , d of cycles in F satisfy 0 { i i} k kα T = (3) 0 β

Proof : Since T0 is the minimal element of the cycle, the next number on the path from T0 will be larger than T0. Therefore, T0 is odd. Given the path is a cycle, Ts = T0, or from Theorem 3,

T0 = Ts 3U T + kα = 0 2U+D kα = 2U+D 3U kα − = β

We have assumed that T0 is the the minimal element of the cycle. That assumption is actually not necessary. We can just assume that T0 is the first odd number after an even number. This formulation will lead to multiple solutions for the same cycle with each orb’s minimum point being a solution. Given each cycle will necessarily have at least one odd and one even element means every cycle will have an odd starting element.

Theorem 6 : There is a cycle in Fk iff there is a solution to equation (3) for some u , d . { i i} Proof : Theorem 5 shows that if there is a cycle, T0 satisfies equation (3). The critical observation is that the formula additionally embeds all the the odd and even conditions of the collatz function. A path from T0 can get back to T0 only if at all iterations, correct odd/even choice is made and all elements on the path are integer. Even a single wrong odd / even path selection results in a fraction. Once a fraction is created, there is no way an integer can result through any number of application of Fk as k is odd. First time a fraction is created from integer through incorrect path selection in Fk will be either applying even path to odd number or applying odd path to even number. a In either case, fraction 2 , where a is an odd integer, results. Subsequent application a of Fk leaves similar fraction 2b , where a is odd, b > 0. Both odd and even paths are

7 considered below: a 3a/2b + k F ( )= k 2b 2 3a + k2b = 2b+1 c = 2b+1 Note that c is odd in above as a is odd and k 2b is even, making the sum odd. As for · even path,

a a/2b F ( )= k 2b 2 a = 2b+1 Theorem 6 has significant implications for collatz conjecture and we capture that in Theorem below.

Theorem 7 : A non-trivial cycle in Collatz sequence exists if and only if for some u , d , β α. { i i} | Proof : From equation (3), the starting point of cycle in collatz sequence will be, α T = . 0 β

Given T is integer, the only way this can happen is if β α. Therefore, there is a non- 0 | trivial cycle if and only β α for some u , d . Alternatively, collatz conjecture will be | { i i} falsified if for any u , d , β α. Obviously, u , d = 1, 1 for trivial cycle satisfies this { i i} | { 1 1} { } equation with T0 = 1.

Theorem 8 : Every cycle u , d in F will inherited by all F where r N. The { i i} k rk ∈ minimum element of the corresponding cycle in Fkr will be rT0, the cycle orbs in Frk be same as u , d and each element in the cycle will be multiplied by r. { i i} Proof : Multiplying the cycle equation (3) for cycle u , d with T starting point, { i i} 0 by r (rk)α r T = · 0 β From Theorem 6, u , d is a cycle in F with starting point rT . It is also evident that { i i} rk 0 all subsequent elements of the cycle in Frk are multiple of r.

Theorem 9 : For every cycle sequence of u , d , u , d , , u , d in F where { 1 1 2 2 · · · s s} k T = kα , cycle sequence u , d , u , d , , u , d , u , d describes the element of the 0 β { 2 2 3 3 · · · s s 1 1} same cycle that succeeds T0 by one orb.

8 Proof : Let the element of a cycle obtained by the sequence u1, d1, u2, d2, , us, ds { · · · }′ be T0 and the element obtained by the sequence u2, d2, u3, d3, , us, ds, u1, d1 be T ′ { · · · } and corresponding α be α . Equation (3),

kα T = 0 β ′ ′ kα T = β

Let the element of the cycle after one orb be T1. From equation (2)

u1 u1 u1 3 T0 + k(3 2 ) T1 = − 2u1+d1 3u1 kα + k(3u1 2u1 ) = β − 2u1+d1 k 3u1 α + β(3u1 2u1 ) = − β 2u1+d1 using simple algebra one can prove that

u u u ′ 3 1 α + β(3 1 2 1 ) α = − 2u1+d1 Therefore,

kα′ T = 1 β ′ T1 = T

′ Therefore, T is the element of the cycle one orb shifted from T0.

Theorem 10 : For every number L, there exists a k such that Fk has more than L cycles. Proof : Using Hardy and Ramanujan asymptotic partition function 1 P (n) eπ√2n/3 ≍ 4n√3 we can find n such that it has more than L rotationally invariant partitions. Set U = D = n, and k = 2U+D 3U . Equation (3) for F reduces to − k

T0 = α.

9 Note that two partitions of U, D yield different values of α. That is because if they yielded same alpha, it would imply same T for two different cycles of u , d , an impossibility. 0 { i i} Therefore, every partition creates a cycle, and every set of rotationally invariant parti- tions create a unique cycle yielding more than L cycles.

There are several consequences and observations based on the above theorems:

For a composite number k = p p p , the cycles of F are union of cycles in all • 1 2 · · · r k the factors of k and the cycles that originate in Fk. For cycles that originate in Fk, k divides β. For example, F175 has 17 non trivial cycles. It inherits 5 cycles from F5, 1 cycle from F7, 2 original cycles of F25, 2 original cycles of F35, and has 7 of its own original cycles. Note that 5, 7 being prime have only original non-trivial cycles and have 5 and 1 cycles, respectively. A cycle inherited in a multiple retains the u , d , whereas all the cycle elements are multiplied. In that sense, u , d { i i} { i i} are invariant across the entire Fk space and appear more fundamental than the cycle elements.

For all non-trivial cycles of F , atleast one of the factor of k divides β and if k is • k prime, k divides β.

For every u , d , such that β > 0, the unique cycle originates in one and only one • { i i} Fk, where β k = , gcd(α, β) 3 β > 0 = D > U log ⇒ 2 2 The unique mapping from u , d to k implies that cycles have a unique ordering. { i i} If k , k are co-primes, F , F do not have any common non trivial cycle. • 1 2 k1 k2 Since 3r does not divide β, all cycles of k = 3r will require β to divide α. That is • the same condition for non-trivial cycle in collatz sequence. This implies that all cycles of F3r are exactly same as cycles of collatz sequence. Collatz conjecture can be proved or disproved for any F3r .

All cycles of F and F p· are same. • k 3 k All elements of path of F (nk) will be multiple of k. Starting with nk, the next • k element in the path

3nk+k 3n+1 2 = k 2 if n 1 mod2 Fk(nk)= ≡ nk = k n if n 0 mod2 ( 2 2 ≡ = k F1(n)

10 For k not divisible by 3 and n not a multiple of k, path of F (n) has no element that • k is multiple of k. This also implies that in such situation Fk(n) will not converge to trivial cycle.

Assuming collatz conjecture to be true, all integers of the form nk converge to • 3p trivial cycle in Fk. Conversely, if number is not of this form, it will not converge to trivial cycle.

nk Assuming collatz conjecture to be true, only and all numbers of the form p con- • 3 verge to the trivial cycle, given an interval with a range that is an integral multiple 3p of k, only 100 k % of the numbers will converge to the trivial cycle where p is the maximum integral value for which 3p divides k.

Under the assumption that F converges, every k that is not a multiple of 3 will • k have atleast one original cycle.

5 Convergence of Generalized Collatz Sequences

Fk(n) is said to converge if there is a repeat number in the path of Fk(n). Fk is said to converge, if it converges for all n > 0. Convergence of Fk for any k will prove the convergence of F1 or the collatz sequence. However, inverse if not true. It is possible that collatz sequence converges but for some k and n, Fk(n) diverges. We can define convergence for Fk(n) as reaching a minimum point of a cycle of Fk. For all odd n (for even n, repeatedly divide the number by 2 to reach an odd number),

p 3U n + kαp T = · 0 2U p+Dp p kαc 3U n + kαp = βc 2U p+Dp

c c c c p p p p where, α , β ,U , D are respective expressions for a cycle of Fk and α , β ,U , D are the expressions for the path from n to T0. The path from n to T0 can be expressed as uc, dc . There is no known solution to the above equation and proving convergence of { i i } GCS remains an open problem. The probabilistic argument used for collatz conjecture is applicable for GCS as well. Assuming equal distribution of odd and even numbers on the Fk paths, and using the heuristic argument as in Lagaris (1985) [7], we can conclude that for sufficiently large n k, the growth between two successive odd number will be 3 forcing the sequence ≫ 4 towards convergence. In our experiments, we saw that average path length to convergence increases with k. Path length to convergence is the number of steps required to complete the cycle starting from the number. For k = 636637, starting with 265797, it took 32253 steps to converge whereas, the average path length to converge for first million numbers was more than twenty thousand steps. Comparatively, average path length to converge for first million numbers for k = 5 is just 53 and maximum path length is 266.

11 Table 1 details the average and maximum path length to convergence for 1 n 106 ≤ ≤ for various k. Column three gives the number that resulted in the maximum path length. The average of the normalized path length for the above n range is also given. Normalized path length σ(n) is defined as

Path length to convergence starting with n σ(n)= ln(n)

Max Max Avg.No. Avg. of Max. Max. Avg.No. Avg. of k Steps Step N Steps σ(n) k Steps Step N Steps σ(n) 1 299 837798 65 3.5 5 266 822266 53 2.9 11 360 959044 60 3.3 19 324 391754 68 3.7 23 325 844124 71 3.9 31 341 842616 76 4.1 39 284 807996 48 2.6 43 312 787220 98 5.3 55 289 853152 61 3.3 59 325 763988 69 3.7 67 345 851708 110 6.0 71 315 917470 76 4.1 79 353 749462 105 5.7 83 274 868268 74 4.0 95 371 899840 94 5.1 101 310 657062 72 3.9 109 405 978932 126 6.8 119 410 513248 101 5.5 121 468 936998 144 7.8 143 512 658640 136 7.4 161 338 835706 83 4.5 169 423 893382 95 5.2 185 446 878182 162 8.8 191 460 492422 127 6.9 199 597 562814 160 8.7 333669 7936 847691 3890 211.9 445419 398 616469 119 6.5 449235 2746 8365 857 46.7 560975 7455 972143 2381 129.7 636637 32253 265797 20227 1101.9 650389 15629 820359 8744 476.3 699075 3915 788597 1693 92.2 787369 15051 431955 8277 451.0 814071 8723 3581 4610 251.1

Table 1: Average Path Length to Convergence for first Million Numbers

We have seen path lengths to convergence running into several hundred thousand. For example, for prime k = 42465127 and n = 3434, the path length to convergence is 202823 steps. It is not hard to think of path length to convergence running into millions for some large k. That said, all Fk(n) do eventually converge. Supported by our exper- iments and heuristic arguments, we state and support the following conjecture from [8]. This still remains unproved.

Extended Collatz Conjecture F (n) converges for all k,n N. k ∈ 6 Number of Cycles in Generalized Collatz Sequences

We showed earlier that F has a trivial cycle for all k. For k = 3p, there is an additional k 6 cycle if there exists a n that is not a multiple of 3 and Fk(n) converges. Non-trivial cycle count for Fk varies. An interesting aspect to explore then is the number of cycles for any Fk. The number of cycles in Fk will be the number of solutions to equation (3). This is an unsolved problem and hence no definitive answer can be provided. However, some insight can be gained into the problem in certain cases. We first explore the case when k is prime. Equation (3) can be rewritten as:

T α = k β 0 · ·

12 Given α, β, T0, and k are all integers, and k is prime, this equation can be satisfied only if:

T is a multiple of k. Trivial cycle k 2k belongs to this category. Given all • 0 → cycles in F1 will be present in Fk, if there is a non-trivial cycle in Fk with T0 = k r, it will be a non-trivial cycle in F1 as well, thus falsifying collatz conjecture. α is a multiple of k. This implies that 2U+D 3U = Rk, where R is any posi- • − tive integer. It is easy to prove that there are infinite solutions to this equation. Equation (3) in this case becomes α T = 0 R Thus, a cycle will result only if R divides α. We cannot prove that for any solution R. Thus, a definitive answer on number of cycle eludes. However, our experiments and all the experiments run to date by other researchers and the heuristic argument suggests that Fk(n) converges to a cycle for all k and n. That implies that in all probability there is atleast one non-trivial cycle in Fk. Infact, for a non-trivial cycle to exist, Fk(n) has to converge to a cycle only for one non k-multiple number. But, that also cannot be proved for all k. Therefore, potentially, there can be none, one, or infinite many non-trivial cycles in Fk. It is also possible that there in no non- trivial cycle for certain Fk, finitely many for some other, and infinite many cycles for others.

There are certain k where we can prove the existence of non trivial cycle in Fk. When there is solution to equation 2U+D 3U = k, there will be cycles in F . In that case, − k equation (3) for FCk reduces to kα T = 0 k = α

That implies each partition of U, D will create a cycle in Fk. { } r There will be non-trivial cycles in Fk, where k is of the form 2 3 for r 3. In r− − − ≥ this case, F (1) = 3+2 3 = 2r 1 which after r 1 steps yield 1 creating a cycle. This k 2 − proves existence of at least one non-trivial cycle in k = 5, 13, 29, .. and their multiples. Similarly, we can consider integer k of the form k = n (4 2r 9)/5. Starting with r · · − n, Fk(n) yields n2 after two up iterations, creating a cycle after r down steps. This yields non-trivial cycle for k = 7, 11, 23, and its multiples. We can keep constructing · · · non-trivial cycles in Fk for various k but it still does not prove that there exits at least one non-trivial cycle for all k.

Some questions to investigate:

Is there an upper bound on the number of cycles for a given F ? • k

13 Are there finite number of cycles in all F ? • k Are there infinite number of cycles in any F ? • k Based on our experiments and heuristic arguments, we make the following conjecture:

Non Trivial cycles in GCF Conjecture For k = 3p, F has at least one but finite number of non trivial cycles. 6 k 7 Integer Space Partition Framework

Assuming Extended Collatz Conjecture to be true, Collatz function (1) along with iter- ation to convergence process can be considered as a integer space partition function.

P : N ζ k → k Since a number can converge to only one cycle, the mapping is exclusive and therefore a partition. Once a partition appears in Fk, it is inherited by all integral multiples of k. Partition e , e , e , ... in F becomes a partition r e , r e , r e , ... in F . For { 1 2 3 } k { 1 2 3 } rk example partitions in F175 include the partitions of F5, F7, F25, F35. The total number of partitions in Fk will be sum of number of original cycles in all the factors of k plus the number of original cycles of Fk. There is infact a stronger inheritance - the entire collatz graph of Fk with all the interconnections between elements is inherited in Frk with elements multiplied by r while the interconnections are retained completely. Thus, collatz graph of Fk consists of all the inherited graphs of all its factors. The collatz conjecture within this framework can be stated as: The collatz function partitions positive integers in only one equivalence class. Table 2 shows the inheritance of known partition set for 1, 5, 7 and 35 . As seen in the table each element of an inherited cycle is an integral multiple of the ele- ment of the original cycle, this property of inheritance of cycles and the partitioning of natural numbers is more fundamental to the conjecture than the path length of a given n.

14 k inherited from/original T0 Equivalence class 1 original 1 1,2,3,4,5 { ···} 1 5 {5 ,10, 15, 20, 25 } · · · 1 1, 2, 4, 8, 9 { ···} 19 3, 6, 7, 11, 12, 13 5 { ···} original 23 23, 29, 37, 46, 51, 58 { ···} 187 123, 187, 246, 251, 283 { ···} 347 171, 259, 342, 347, 359 { ···} 1 7 {7, 14, 21, 28, 35 7 ···} original 5 1, 2, 3, 4, 5,6 ,8 ,9 { ···} 1 35 35, 70, 105, 140, 175, 210 { ···} 7 7, 14, 28, 56, 63, 112 { ···} 133 21, 42, 49, 77, 84, 91 { ···} 5 161 161, 203, 259, 322, 357, 406 { ···} 35 1309 861, 1309, 1722, 1757, 1981, 2037 { ···} 2429 1197, 1813, 2394, 2429, 2513, 2737 { ···} 7 25 5, 10, 15, 20, 25, 30 { ···} 13 1, 2, 4, 8, 9, 13 original { ···} 17 3, 6, 11, 12, 17, 22 { ···} Table 2: Inheritance seen in cycles of 5,7 and 37

8 Experimental Results

We generated GCS for all k < 200 and some randomly selected large k for all n up to 1 billion. The catalog of all non-trivial cycles found for a few numbers are in Table 3. Table gives original cycles, total cycles, the factors from where cycles are inherited, and the cycle starting element of the original cycles. An interesting question we explored is the percentage of numbers converging to various cycles of Fk. Our experiments did not show any pattern in numbers converging to a particular cycle for prime k. For composite k, one can predict the numbers converging to cycles inherited from various factors. We detail these finding using cycles of F and F = 11 17. Each number is 5 187 × mapped to the T0 of the cycle it converges. For k = 5, we compute the distribution of the numbers converging to various cycles by dividing 100 million into 200 intervals of 500, 000 each. For each interval, the percentage of numbers converging to various cycles is computed. The distribution is shown in Figure 1. Recall there are six known cycles in F with ζ 1, 5, 19, 23, 187, 347 . Every 5th number converges to trivial cycle and 5 5 ⊇ { } hence 20% of numbers converge to trivial cycle T0 = 5. The percentage distribution for various cycles in different buckets seems random and unpredictable. It almost appear to diverge. Perhaps a machine learning program may be able to find some pattern in the distribution.

15 In case of composite number 187 = 11 17, F inherits two cycles from F , two × 187 11 from F17, trivial cycle from F1, and there are two cycles that originate in F187. No new information emerges in looking at the distribution of numbers among these cycles. However, we can predict the percentage of numbers converging to the four cycle groups - cycles inherited from k = 11, cycles inherited from k = 17, cycles inherited from k = 1, and original cycles. Every multiple of 187 converges to trivial cycle. Every multiple of 11 barring multiple of 187 converges to cycles inherited from F17, every multiple of 17 barring multiple of 187 converges to cycles inherited from F11. All other numbers converge to cycles originating in F187. That is why figure 2 shows a flat distribution for all four groups. Here, 187 200 was divided into 200 intervals of 187 size each. ∗

16 Original Total Inherited k Cycles Cycles From T0 of Original Cycles 5 5 5 1, 19, 23, 187, 347 7 1 1 5 11 2 2 1, 13 13 9 9 1, 131, 211, 259, 227, 287, 251, 283, 319 17 2 2 1, 23 23 3 3 5, 7, 41 25 2 7 5 7, 17 29 4 4 1, 11, 3811, 7055 35 2 8 5, 7 13, 17 37 3 3 19, 23, 29 43 1 1 1 47 7 7 25, 5, 65, 89, 73, 85, 101 53 1 1 103 61 2 2 1, 235 77 1 4 7, 11 1 79 4 4 1, 7, 233, 265 89 1 1 17 95 3 9 5, 19 1, 23, 17 97 2 2 1, 13 101 7 7 11, 29, 7, 19, 23, 31, 37 103 2 2 23, 5 115 2 10 5, 23 13, 17 119 5 8 7, 17 1, 5, 11, 23, 125 121 2 4 11 5, 19 127 2 2 1, 41 131 3 3 13, 23, 17 139 1 1 11 145 3 12 5, 29 1, 47, 23 149 2 2 19, 667 155 1 7 5 31 1 157 1 1 13 169 2 11 13 11, 17 173 2 2 7, 37 181 3 3 23, 55, 11 185 1 9 5, 37 1 199 2 2 13, 47

Table 3: New Cycles and inherited Cycles for k

17 17 21.00

16 20.75

20.50 15 20.25

14 20.00

19.75 13

19.50 12 percentage of numbers mapped mapped numbers of percentage percentage of numbers mapped mapped numbers of percentage 19.25

11 19.00 0 25 50 75 100 125 150 175 200 0 25 50 75 100 125 150 175 200 Interval Interval

T0 = 1 T0 = 5

54 13

12 52 11

50 10

9 48 8

percentage of numbers mapped mapped numbers of percentage 46 mapped numbers of percentage 7

6 0 25 50 75 100 125 150 175 200 0 25 50 75 100 125 150 175 200 Interval Interval

T0 = 19 T0 = 23

6 5.0

5 4.5

4.0 4

3.5 3

3.0 percentage of numbers mapped mapped numbers of percentage percentage of numbers mapped mapped numbers of percentage 2

2.5 0 25 50 75 100 125 150 175 200 0 25 50 75 100 125 150 175 200 Interval Interval

T0 = 187 T0 = 347

Figure 1: Percentage of numbers converging to various cycles of k = 5. Interval is 500k

18 9.0 5.6

8.8 5.5

5.4 8.6

5.3

8.4

5.2 percentage of numbers mapped numbers of percentage mapped numbers of percentage 8.2 5.1

0 25 50 75 100 125 150 175 200 0 25 50 75 100 125 150 175 200 Interval Interval Cycles inherited from k = 17 Cycles inherited from k = 11

90

0.56

88 0.55

0.54 86

0.53 84

0.52 percentage of numbers mapped 187 mapped numbers of percentage percentage of numbers mapped numbers of percentage 82

0.51 0 25 50 75 100 125 150 175 200 0 25 50 75 100 125 150 175 200 Interval Interval Cycles inherited from k = 1 Cycles Originating in k = 187

Figure 2: Percentage of numbers converging to various cycles of k = 187. Interval is 187

Given in Table 4 are u , d for all cycles of 5 and 51. { i i} k T Tot. steps u d 0 { i} { i} 5 1 3 1 2 { } { } 5 19 5 3 2 { } { } 5 5 2 1 1 { } { } 5 23 5 2,1 1,1 { } { } 5 187 27 6,3,2,1,1,4 1,1,1,2,1,4 { } { } 5 347 27 5,5,1,1,2,2,1 2,1,1,3,1,1,1 { } { } 51 69 31 3,2,1,4,1,1,3,2,1 1,1,2,1,1,1,3,2,1 { } { } 51 3 7 1,1 1,4 { } { } 51 51 2 1 1 { } { }

Table 4: list of ui, di for each cycles found in F51, F5

To illustrate the point that every sequence of u , d creates a cycle, we created some { i i} random u , d sequences to find the F where the cycle originated. A random number { i i} k between 5 and 15 was chosen to determine the orb count. Subsequently, numbers between

19 1 and 3 were chosen randomly to create a sequence of u ,d and put in equation 3 to { i i} compute the originating k for the cycle.

orbs ui di T0 k { } { } 14 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 1, 2, 40007869221581 34901942552351 {2, 2, 1, 1, 1, 2, 1 {2, 2, 1, 1, 2, 1, 1 } } 10 2, 2, 1, 1, 1, 1, 2, 2, 3, 2, 42639161743 68590336573 {3, 1, 3, 1, 2 {2, 2, 1, 1, 3 } } 6 3, 1, 1, 3, 2, 3, 66984883 134040581 1,{ 3, 2 3,{ 3, 2 } } 9 3, 1, 1, 2, 3, 2, 3, 1, 2, 1, 183478133657 30872953967 { 3, 2, 2, 3 { 2, 2, 1, 1 } } 14 3, 2, 2, 2, 1, 1, 1, 3, 3, 1, 3, 1, 30452051799122219 36027949730354525 3,{ 2, 1, 2, 1, 2, 1, 2 2,{ 2, 2, 3, 3, 3, 2, 1 } } 8 1, 2, 2, 1, 1, 1, 2, 2, 17567383 16245775 {2, 1, 1, 2 {2, 1, 2, 1 } }

Table 5: Cycles with originating k and T0 for random ui,di

Another interesting aspect of the GCS cycles is that almost all of them originate at values less than k as shown in table 6. We took 100 random values of k and found all the cycles that originated below 100 million, we then took the maximum value in ζk and divided it by k, most of the values were less than 0.1, of those that were greater than 0.1, the cycles had been reached by an n less than k. This aspect is also evident in cycle data in table 3. This does not proof that cycles originating at higher number may not exist but the empirical evidence suggests that they may not.

20 k cyc Max T0 Max T0/k k cyc Max T0 Max T0/k k cyc Max T0 Max T0/k 11089 2 59 0.01 46429 1 35 0 74081 3 4825 0.07 13501 2 175 0.01 47821 13 3053 0.06 74287 1 637 0.01 15095 11 1387 0.09 48293 1 221 0 75583 2 343 0 15677 3 389 0.02 49811 1 65 0 76643 1 55 0 15823 2 881 0.06 50183 4 2215 0.04 77443 16 7939 0.1 16133 9 641 0.04 51253 2 37 0 77773 1 871 0.01 16189 1 703 0.04 52159 2 961 0.02 79861 1 55 0 17177 2 199 0.01 52225 5 4363 0.08 80029 2 515 0.01 19529 1 7 0 52741 24 7079 0.13 80149 5 1475 0.02 19753 2 361 0.02 53333 3 527 0.01 80441 2 359 0 20615 23 7327 0.36 56461 23 8281 0.15 81607 7 7021 0.09 20777 3 295 0.01 56827 1 1237 0.02 82571 5 2027 0.02 20801 40 5393 0.26 57217 8 419327 7.33 83801 3 1775 0.02 22343 2 53 0 57467 4 17785 0.31 83875 40 12989 0.15 24211 46 6203 0.26 59485 1 431 0.01 84827 1 151 0 25247 4 227 0.01 61349 4 1279 0.02 85945 12 6233 0.07 25399 2 263 0.01 61457 13 1871 0.03 86431 1 35 0 25711 5 871 0.03 62605 4 839 0.01 86711 4 1373 0.02 31313 2 119 0 62873 1 481 0.01 87347 2 881 0.01 31727 136 328241 10.35 67745 7 1661 0.02 87389 84 21071 0.24 34241 7 14939 0.44 68969 46 14215 0.21 87461 1 101 0 34985 2 503 0.01 69193 54 16129 0.23 87475 3 2143 0.02 35281 13 911 0.03 69349 1 43 0 87749 6 1405 0.02 36143 7 1301 0.04 69593 1 451 0.01 88915 3 6491 0.07 37109 3 1691 0.05 69733 1 5 0 88997 1 869 0.01 37669 2 169 0 69745 27 7417 0.11 91229 1 55 0 37843 9 4819 0.13 69961 11 4163 0.06 92453 1 629 0.01 37925 9 1919 0.05 70181 10 2305 0.03 92573 5 1409 0.02 38545 12 2867 0.07 71099 1 463 0.01 93007 2 607 0.01 39029 1 1459 0.04 71267 1 25 0 93335 2 739 0.01 39815 7 1561 0.04 72949 1 19 0 94835 3 893 0.01 43081 6 5791 0.13 72955 2 329 0 98021 26 9707 0.1 43661 1 287 0.01 73169 1 1 0 98167 5 3425 0.03 44381 53 46111 1.04

Table 6: The number of original cycles and maximum T0 for random k

9 Relationship between GCS and Diophantine Equation

Cycles of F offer solution to the Diophantine equation 2m 3n = k. Algorithm to solve k − the equation is below Algorithm to Solve 2m 3n = k − Input k Output m,n Step 1 : Set r = 1 Step 2 : Run Fk(r) till convergence. Step 3 : Compute M = 2U+D 3U for the converged cycle. − Step 4 : If M = k, solution found. Output U + D,U. Stop. else increment r by 2, goto Step 2.

The above algorithm does not stop if there is no solution to the Diophantine equation. We are working on the conditions to when algorithm should stop with a failure. An area of further research is to get insight into why collatz sequence generation results in solution to the Diophantine equation.

21 10 Collatz Conjecture

Using equation (2), convergence of collatz sequence can be expressed as - for all n, does there exist u , d , such that { i i} 3U n + α 1= 2U+D 2U+D = 3U n + α

There is no known solution to the above equation for every number and the research continues.

11 Conclusion

We examined the cycles of GCS and in particular focused on convergence and inheritance of these cycles across F . An insight is orbs u , d are invariant across F space and k { i i} k appear more fundamental than the cycle elements. It is their interrelationship that creates a cycle in the appropriate Fk and in some sense they behave somewhat similar to Pythagorean triplets. An area of further research is to explore properties of Pythagorean triplets that may have applicability towards GCS cycles. As is expected and conjectured, F is convergent for all k and there is atleast one non-trivial cycle for F , k = 3p. k k 6 References

[1] Edward G. Belaga, Maurice Mignotte. Embedding the 3x+1 Conjecture in a 3x+d Context Experimental Mathematica 7:2, 145-151.

[2] Eliahou, Shalom The 3x + 1 problem: new lower bounds on nontrivial cycle lengths Discrete Mathematics. 118 (1): 45–56. doi:10.1016/0012-365X(93)90052-U.

[3] Eric Roosendaal. On the 3x + 1 problem http://www.ericr.nl/wondrous/

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[5] John Lesieutre. On a Generalization of the Collatz Conjecture https://www.semanticscholar.org/paper/On-a-Generalization-of-the-Collatz- Conjecture-John-Wang/9da2c36caa6a8aca9ed0c3a03b3fe0efd32eb2a8

[6] Kurtz, Stuart A.; Simon, Janos (2007). The Undecidability of the Generalized Col- latz Problem In Cai, J.-Y.; Cooper, S. B.; Zhu, H. (eds.). Proceedings of the 4th International Conference on Theory and Applications of Models of Computation, TAMC 2007, held in Shanghai, China in May 2007. pp. 542–553. doi:10.1007/978-3- 540-72504-6-49. ISBN 978-3-540-72503-9. As PDF

22 [7] Lagarias, Jeffrey C. The 3x + 1 problem and its generalizations The American Mathematical Monthly. 92 (1): 3–23. doi:10.1080/00029890.1985.11971528. JSTOR 2322189.

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[9] Lagarias, Jeffrey C. The ultimate challenge: the 3x + 1 problem Providence, R.I.: American Mathematical Society. p. 4. ISBN 978-0821849408.

[10] Simons, John L. On the nonexistence of 2-cycles for the 3x + 1 problem Math. Comp. 74: 1565–72. Bibcode:2005MaCom..74.1565S. doi:10.1090/s0025-5718-04- 01728-4. MR 2137019.

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[13] Tao, Terence. Almost all Collatz orbits attain almost bounded values https://arxiv.org/abs/1909.03562

23 0.56

0.55

0.54

0.53

0.52 percentage of numbers mapped numbers of percentage

0.51

0 25 50 75 100 125 150 175 200 Interval